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Difference between revisions of "2014 AMC 10B Problems/Problem 21"

(→‎Problem)
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[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
  
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==Solution==
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==Solution 1==
  
 
<asy>
 
<asy>
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The two diagonals are <math>\overline{AC}</math> and <math>\overline{BD}</math>. Using the Pythagorean theorem again on <math>\bigtriangleup AFC</math> and <math>\bigtriangleup BED</math>, we can find these lengths to be <math>\sqrt{96+529} = 25</math> and <math>\sqrt{96+961} = \sqrt{1057}</math>. Since <math>\sqrt{96+529}<\sqrt{96+961}</math>, <math>25</math> is the shorter length, so the answer is <math>\boxed{\textbf{(B) }25}</math>.
 
The two diagonals are <math>\overline{AC}</math> and <math>\overline{BD}</math>. Using the Pythagorean theorem again on <math>\bigtriangleup AFC</math> and <math>\bigtriangleup BED</math>, we can find these lengths to be <math>\sqrt{96+529} = 25</math> and <math>\sqrt{96+961} = \sqrt{1057}</math>. Since <math>\sqrt{96+529}<\sqrt{96+961}</math>, <math>25</math> is the shorter length, so the answer is <math>\boxed{\textbf{(B) }25}</math>.
 +
 +
===Solution 2===
 +
 +
<asy>
 +
size(7cm);
 +
pair A,B,C,D,CC,DD;
 +
A = (-2,7);
 +
B = (14,7);
 +
C = (10,0);
 +
D = (0,0);
 +
E = (4,7);
 +
draw(A--B--C--D--cycle);
 +
draw(D--E);
 +
label("10",(A+D)/2,W);
 +
label("14",(B+C)/2,E);
 +
label("$A$",A,NW);
 +
label("$B$",B,NE);
 +
label("$C$",C,SE);
 +
label("$D$",D,SW);
 +
label("$D$",D,SW);
 +
label("$21$",(C+D)/2,S);
 +
</asy>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=20|num-a=22}}
 
{{AMC10 box|year=2014|ab=B|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:35, 19 January 2019

Problem

Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length $33$ and $\overline {CD}$ of length $21$. The other two sides are of lengths $10$ and $14$. The angles $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD$?

$\textbf{(A) }10\sqrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qquad\textbf{(D) }18\sqrt{2}\qquad\textbf{(E) }26$

Solution 1

[asy] size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label("33",(A+B)/2,N); label("21",(C+D)/2,S); label("10",(A+D)/2,W); label("14",(B+C)/2,E); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",DD,N); label("$F$",CC,N); draw(C--CC); draw(D--DD); [/asy]

In the diagram, $\overline{DE} \perp \overline{AB}, \overline{FC} \perp \overline{AB}$. Denote $\overline{AE} = x$ and $\overline{DE} = h$. In right triangle $AED$, we have from the Pythagorean theorem: $x^2+h^2=100$. Note that since $EF = DC$, we have $BF = 33-DC-x = 12-x$. Using the Pythagorean theorem in right triangle $BFC$, we have $(12-x)^2 + h^2 = 196$.


We isolate the $h^2$ term in both equations, getting $h^2= 100-x^2$ and $h^2 = 196-(12-x)^2$.

Setting these equal, we have $100-x^2 = 196 - 144 + 24x -x^2 \implies 24x = 48 \implies x = 2$. Now, we can determine that $h^2 = 100-4 \implies h = \sqrt{96}$.

[asy] size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label("33",(A+B)/2,N); label("21",(C+D)/2,S); label("10",(A+D)/2,W); label("14",(B+C)/2,E); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$D$",D,SW); label("$E$",DD,SE); label("$F$",CC,SW); draw(C--CC); draw(D--DD); label("21",(CC+DD)/2,N); label("$2$",(A+DD)/2,N); label("$10$",(CC+B)/2,N); label("$\sqrt{96}$",(C+CC)/2,W); label("$\sqrt{96}$",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]

The two diagonals are $\overline{AC}$ and $\overline{BD}$. Using the Pythagorean theorem again on $\bigtriangleup AFC$ and $\bigtriangleup BED$, we can find these lengths to be $\sqrt{96+529} = 25$ and $\sqrt{96+961} = \sqrt{1057}$. Since $\sqrt{96+529}<\sqrt{96+961}$, $25$ is the shorter length, so the answer is $\boxed{\textbf{(B) }25}$.

Solution 2

size(7cm);
pair A,B,C,D,CC,DD;
A = (-2,7);
B = (14,7);
C = (10,0);
D = (0,0);
E = (4,7);
draw(A--B--C--D--cycle);
draw(D--E);
label("10",(A+D)/2,W);
label("14",(B+C)/2,E);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$D$",D,SW);
label("$21$",(C+D)/2,S);
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See Also

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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